Stability of pde in some $L^p$ norm and stability of a numerical scheme for it equivalence.

I would like to get some light on how to proceed and my confusion. I consider some IBVP of the form $$u_t+L(t,x)u=0, x\in D, t\in [0,T]$$ with some BC and initial data. And I use some numerical method to solve it. First, I start with a question of well-posedness of IBVP. Assume I proved existence and uniqueness in some general space, say only bounded measurable functions, so in fact I showed existence and uniqueness only of a weak solution. Now, I still have to show stability and I can see it perfectly overlaps with a question of stability of a numerical scheme. And this is where I am stuck. So, I can pick a space first, and as it is done in many books, I choose $L^2$ and try to show stability, that is to prove that $$||u(t,x)||_2 \leq C ||u(0,x)||_2$$ for some $C$ using energy methods, for example. I can also use discrete energy methods and discrete energy norms to show the same for a numerical method that $$||u_h(t_n,x)||_2 \leq C_h ||u_h(0,x)||_2.$$ I am dealing with a problem where I have these stability results for any $t,t_n \in [0,T-\epsilon]$, with $C=C(\epsilon)$, but not on $[0,T]$. If I picked $L^{\infty}$ instead of $L^2$, I can show it for $t\in [0,T]$ and $C$ is a fixed constant. So I have the following questions:

1. So far, looks like a problem is well posed only in $L^{\infty}$ on $[0,T]$. However, I happen to note that my numerical method does exhibit $L^2$ convergence, so it might be a coincidence or there is some limiting result I can show in $L^2$ in this case?

2. What is the intuition behind choosing $L^2$ or $L^{\infty}$(I can also pick some Sobolev space but I know nothing about them, so I restrict myself to $L^p$ spaces). Is one better than the other? Is it stronger result in one space? I know that $L^{\infty} \in L^2 \in L^1$ but there is no inequality between those norms though, so I can't say that stability in one norm imply stability in another, can I?

3. Related to 2. So, If I have a stable IBVP in $L^p$ my numerical scheme has to be stable at least in that space? Say I have $L^2$ well posed problem, can I try to prove $L^1$ stability of a numerical scheme or it doesn't make sense since the initial problem is not $L^1$ stable?

-

1. Stability in $L^\infty$ should imply $L^2$-stability (if your domain is bounded) since you have the bound $\|\cdot\|_{L^2}\leq C\|\cdot\|_{L^\infty}$.
2. You have the continuous embedding $L^p(U)\subset L^q(U)$ for $p>q$, provided $U$ is bounded. This means that $\|u\|_{L^q}\leq C\|u\|_{L^p}$ for $u\in L^p(U)$.
 thanks! for the 1.: when you say "imply", you mean that if I have (1)$||u(t)||_2\leq||u(0)||_2$ I can only deduce (2)$||u(t)||_2\leq||u(0)||_{\infty}$, but not that (3)$||u(t)||_{\infty}\leq||u(0)||_{\infty}$. Both (1) and (2) imply stability as the norm in $L^2$ is bounded, just with different constants? Is (1) a sharper bound? second, I agree that I have these relation between $L^2$ but in discrete all norm equivalences break down. Do your arguments hold in the discrete case as well as the mesh approaches zero? – Medan Jan 16 at 1:44 @Medan: What I meant is that if the error is small in $L^\infty$, it must be small also in $L^2$, because of this relation. For the second question, you must use a continuous norm to measure the error, or a discrete norm that behaves like the continuous version of it. Otherwise it does not really make sense to talk about convergence. – timur Jan 16 at 3:37 I agree that $L_{\infty}$ boundedness would imply boundedness of $L_2$ norm, but I am asking if in continuous world $||\cdot||_{\infty}\leq||\cdot||_{\infty}$. Perhaps not, thus I am asking if $||\cdot||_2\leq ||\cdot||_{\infty}$ is still equivalent to say it is stable in $L_2$ but not in $L_{\infty}$. – Medan Jan 16 at 4:06 And for the second question I am using a discrete version, say $||u||^2=h\sum u^2_i$. However, I remember the fact that I can't rely in finite difference world on any relationship between the norms as the constants might blow up in the limit. Thus, if using the discrete norms for $L_2$ and $L_{\infty}$ I am able to show only $||\cdot||_2\leq ||\cdot||_{\infty}$ for the method, would that still imply my numerical method is stable? – Medan Jan 16 at 4:10 @Medan: I think I wrote stability instead of convergence in my answer. Please read "convergence" instead of "stability". Your first question was about convergence. For the second question, not problem, because your discrete norm is designed to behave like the continuous $L^2$-norm. Constants will blow up only if you were using a wrong norm. – timur Jan 16 at 4:37